The Erasmus pentagon

Question

Professor Erasmus has constructed a special convex pentagon $ABCDE$ that he modestly calls the "Professor-Erasmus-pentagon". The professor claims that he can cut off a smaller pentagon similar to pentagon $ABCDE$ by a straight line.

Question: Has the professor once again made one of his well-known mathematical blunders, or do such pentagons indeed exist?

_{Note: The five points A,B,C,D,E are all distinct. There are no tricks (like paper folding; pentagons made of rubber; etc).}

Answer 1

Let $A=(-30,60),B=(0,0),C=(150,0),D=(300,300),E=(225,400)$. $F=(-12,84)$ is on $AE$ and $G=(175,50)$ is on $CD$. Then $ABCDE$ is similar to $FABCG$, by a factor of $\sqrt5$.

The key point of this solution is that angles $GFA, EAB, ABC, BCD, CDE$ are all equal ($\arctan-2\approx116.57^\circ$) and $FA,AB,BC,CD$ are in geometric sequence. Then, we can rotate the pentagon so that each of those edges corresponds to the next one in the smaller pentagon.

Image of the pentagon:

The same idea can be done with other values of the key angle between $90^\circ$ and $135^\circ$ and other values for the ratio of similarity.

Answer 2

Intuitively, I would make a polygon with sides $AB$, $BC$, $CD$, $DE$, $EA$ having lengths in increasing order $1,a,a^2,a^3,a^4$ such that the distance between, for some $a > 1$.

Then, put a point $E'$ such that $E'A = a^{-1}$ and cut the polygon on the line $E'D$. We just have to make sure that $E'D = ED = a^4$. If so, all sides will be $a^{-1},1,a,a^2,a^3$.

Edit: This is how the picture would look like (approximately, don't have any proper tool for drawing). All angles except the one between $a^3$ and $a^4$ are equal.

Answer 3

This is not possible.

Proof:

Look at angle $C$ and find the new angle it corresponds to in the smaller pentagon. If it is one of the other angles of the larger pentagon, then every angle is congruent, so it is a regular pentagon, and this case is trivially impossible.

If $C$ and $C'$ are in the same place, then $AE=A'E'$, but the second pentagon must be smaller than the first. So $C$ must correspond to one of the newly created angles.

Of course, this applies to every angle, not just $C$. So every angle on the pentagon must either correspond to itself or a newly-created angle. Only 2 new angles can be created in any convex pentagon made from a cut through a convex pentagon. Every angle must stay in the exact same place, so the only possible solution is one with no cut at all.

Edit: This has been invalidated by f'''s example - having four congruent angles is possible in this context.

Answer 4

This is possible. The trick is to

Create an asymmetric Pentagon, which can be "flipped" by one cut. Use that similar allows to mirror an object.

My solution looks like this:

I know that

My image isn't exact, since the ratios of the sides need to be the same (e.g. ED/CB = CB/DX).